You have found the following ages (in years) of all 4 gorillas at your local zoo: $ 7,\enspace 10,\enspace 15,\enspace 21$ What is the average age of the gorillas at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 10 + 15 + 21}{{4}} = {13.3\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-6.3$ years $39.69$ years $^2$ $10$ years $-3.3$ years $10.89$ years $^2$ $15$ years $1.7$ years $2.89$ years $^2$ $21$ years $7.7$ years $59.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{39.69} + {10.89} + {2.89} + {59.29}} {{4}} $ $ {\sigma^2} = \dfrac{{112.76}}{{4}} = {28.19\text{ years}^2} $ The average gorilla at the zoo is 13.3 years old. The population variance is 28.19 years $^2$.